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View Full Version : Odds of set-over-set on flop?

10-02-2005, 01:06 AM
If you are dealt a pocket-pair in holdem, the odds of you flopping a set or better are 7.5/1.

What are the odds of another player also flopping a set? My first thought was that it would be (7.5*7.5)/1 or 56/1 but I think it's actually smaller than that since there are only 2 cards left to make the set (although there are different combinations of ways the 2 sets can be made)

Also, I realize this can be a more complicated problem, since there are the odds of another player having a pocket pair to begin with, and then the odds of that player also hitting a set on the flop.

Odds of one player getting set:
50 unknown cards, 2 make the set
odds of not making set: 48/50 * 47/49 * 46/48
odds of making set is 1 - above = .118 or 1/8.5

Forgetting the odds of getting the pocket pairs preflop, assuming two players see the flop and both have pocket pairs, how would you compute the odds of them BOTH hitting sets?

BruceZ
10-02-2005, 01:26 AM
[ QUOTE ]
Forgetting the odds of getting the pocket pairs preflop, assuming two players see the flop and both have pocket pairs, how would you compute the odds of them BOTH hitting sets?

[/ QUOTE ]

2*2*44/C(48,3) =~ 1.02% =~ 97.3-to-1.

KJL
10-02-2005, 01:34 AM
Well since the odds of flopping a set for one player is .11 the odds of 2 doing it would be about .11*.11=.012, but this isn't exactly correct because the second player has slightly better odds. So the correct calculation would be:
C(2,1)*C(2,1)*C(46,1)/C(48,3)=.106 or ~1.1%

Note: that calculation includes the chance that one one person flops a set and one flops quads, if you didn't want that you can replace the number 46 with 44, but the answer would still be the same.

BruceZ
10-02-2005, 01:46 AM
[ QUOTE ]
Well since the odds of flopping a set for one player is .11 the odds of 2 doing it would be about .11*.11=.012, but this isn't exactly correct because the second player has slightly better odds. So the correct calculation would be:
C(2,1)*C(2,1)*C(46,1)/C(48,3)=.106 or ~1.1%

Note: that calculation includes the chance that one one person flops a set and one flops quads, if you didn't want that you can replace the number 46 with 44, but the answer would still be the same.

[/ QUOTE ]

You actually double counted each case where one player flops quads and one flops a set. There are only 4 flops that give one player quads and the other player a set. To include these, the expression is (2*2*44 + 4)/C(48,3) =~ 1.04%.

10-02-2005, 01:57 AM
Forgive the newbie question (it's been a long time since I took statistics) but can you break down what each part of the formula is? Is it 2 players, 2 outs (each), 44 cards that don't help, and the # combinations of 3 cards out of 48 remaining?

Luzion
10-02-2005, 02:03 AM
He is using hypergeometric formula but broke it down for you. The entire formula would be

[ C(2,1) * C(2,1) * C(44,1) ] / C(48,3)

C(2,1) means you hit your set using exactly 1 card out of possible 2 outs. The other C(2,1) means your opponent hits his set using exactly 1 card out of his possible 2 outs. C(44,1) means a safe card that wont give you guys quads, divided by C(48,3) which is all possible combinations of cards that can flop.

BruceZ
10-02-2005, 02:05 AM
[ QUOTE ]
Forgive the newbie question (it's been a long time since I took statistics) but can you break down what each part of the formula is? Is it 2 players, 2 outs (each), 44 cards that don't help, and the # combinations of 3 cards out of 48 remaining?

[/ QUOTE ]

2 cards that make one set, times 2 cards that make the other set, times 44 cards for the 3rd card, divided by # combinations of 3 cards out of 48 remaining.

Or just use fractions:

2/48 * 2/47 * 44/46 * 6.

We multiply by 6 since the cards can come in any of 6 orders. Note that this is the same as 2*2*44/C(48,3) since C(48,3) = 48*47*46/6.

10-02-2005, 02:18 AM
Ah, that makes sense. Thanks a lot!